Lecture 16 - Types of Transformations

Recall that an onto (surjective) means that all outputs have a corresponding input. We also have one-to-one (injective) that implies that any outputs that are the same have the same inputs. A bijective function is both injective and surjective.

An injective

T

$T \in L (V, W)$ is injective iff $null (T) = {\vec{0}}$ .

Proof
Consider $\to$ . Suppose that $T$ is injective. We need to show that $null (T)$ is just the zero vector. Let $x \in null (T)$ . So then $T (x) = \vec{0}$ . Notice that also $T (\vec{0}) = \vec{0}$ . Since $T$ is an injective map, then since $T (x) = T (\vec{0})$ then $x = \vec{0}$ .

Consider $\leftarrow$ . Suppose that $null (T) = {\vec{0}}$ . Let $T (x) = T (y)$ where $x, y \in V$ are arbitrary. Then we can say:

T (x) - T (y) = \vec{0} = T (x - y)

So since since the nullspace only contains the zero vector, $x - y = \vec{0}$ so $x = y$ finishing showing injectivity.
☐
Recall that:

dim (V) = dim (null (T)) + dim (range (T))

Note that if $dim V = n$ and $dim W = m$ :

If $n < m$ , then $T$ cannot be onto (surjective). If the dimension of the domain is less than that of the co-domain, then there's no way to map all the $n$ values to some configuration of $m$ values.
If $n > m$ , then we get the opposite effect, where $T$ is not injective.
If $n = m$ then $T$ is injective if and only if $T$ is surjective.

Getting into 3.C: Matrices

Recall that linear transformations are defined by their actions on the basis vectors.

Suppose we have:

Where $B = (v_{1}, . . ., v_{n})$ is a basis for $V$ , and $C = (w_{1}, . . ., w_{m})$ . We need to write down all the transformations as:

\begin{aligned} T (v_{1}) & = A_{11} w_{1} + A_{21} w_{2} + \dots + A_{m 1} w_{m} \\ T (v_{2}) & = A_{12} w_{1} + A_{22} w_{2} + \dots + A_{m 2} w_{m} \\ ⋮ \\ T (v_{n}) & + A_{1 n} w_{1} + A_{2 n} w_{2} + \dots + A_{m n} w_{m} \end{aligned}

So if you know all the $A_{i j}$ 's then you would have the whole picture of the transformation. So then you can just extract these coefficients and put them into an array called a matrix:

A = [\begin{matrix} A_{11} & A_{12} & \dots & A_{1 n} \\ A_{21} & A_{22} & \dots & A_{2 n} \\ ⋮ & ⋱ \\ A_{m 1} & A_{m 2} & \dots & A_{m n} \end{matrix}]

This is defined to be $M (T, B, C)$ , which is called the matrix representation of $T$ w.r.t. the bases $B$ and $C$ .

The flipped diagonal

The reason for flipping the $A_{i j}$ 's is purely convention at the moment. We'll see later that the matrix arithmetic makes more sense in this notation, so trust us for a moment.

Let's do an example. Let's consider the derivative map $D : P_{4} (R) \to P_{3} (R)$ . Let $B = {1, x, x^{2}, x^{3}, x^{4}}$ and $C = {1, x, x^{2}, x^{3}}$ , so the standard bases for each of their respective sets. The definition above implies we should calculate the transformations into vectors of our $W$ :

\begin{aligned} D (1) & = 0 + 0 x + 0 x^{2} + 0 x^{3} \\ D (x) & = 1 + 0 x + 0 x^{2} + 0 x^{3} \\ D (x^{2}) & = 0 + 2 x + 0 x^{2} + 0 x^{3} \\ D (x^{3}) & = 0 + 0 x + 3 x^{2} + 0 x^{3} \\ D (x^{4}) & = 0 + 0 x + 0 x^{2} + 4 x^{3} \end{aligned}

So then our matrix is:

M (D, B, C) = [\begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \end{matrix}]

Notice here in general, if the $dim (V) = n$ and $dim (W) = m$ then we get an $M_{m \times n}$ matrix.

Food for Thought

What if we knew that $T (p (x)) = p^{'} (x) + p (x)$ ? Notice that $T$ is still a linear map, and supposing that $T : P_{3} (R) \to P_{3} (R)$ . We know it's a $M_{4 \times 4}$ matrix, and we'll get:

M_{4 \times 4} = [\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{matrix}]

Which notice that:

M (4 \times 4) = M (D, B, C) + I_{4}

This implies the following properties, where $T_{1}, T_{2} \in L (V, W)$ :

$M (T_{1} + T_{2}, B, C) = M (T_{1}, B, C) + M (T_{2}, B, C)$
$M (α T_{1}, B, C) = α M (T, B, C)$

So $M$ is a linear map!!! It's a linear map from $L (V, W) \to F^{dim W, dim V}$ .